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I like to know when the multiplication of a matrix by it's transpose will be equal to 1(I) and when it will be equal to 0?

Thanks in advanced!

user10354138
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user3486308
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    What are your thoughts on the problem? Have you been able to find some specific examples? Please put your effort in the question via an edit. – YiFan Tey Oct 27 '18 at 10:37
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    Are the matrices real? –  Oct 27 '18 at 10:46
  • @user496634: I just saw in some calculations they replaced A * A^T =1 and wanted to know is there any situations this will happen always? I searched after I sent my question here, and found that one of the situations would be for orthogonal matrices. So now I like to know are there any more or not? – user3486308 Oct 27 '18 at 10:47
  • @SaucyO'Path: Yes, but I like to know all situations. – user3486308 Oct 27 '18 at 10:48

3 Answers3

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Suppose we assume that $A$ is a real matrix. Then if $A$ is a square matrix and $AA^T=I$, then $A^{-1}=A^T$, which is true for orthogonal matrices. If $AA^T=0$, then $A$ has to be singular since $\det A=0$ (can you see why?) and any two rows of $A$ has to be orthogonal; in particular, any row of $A$ is orthogonal to itself, so the length of that vector in any row is $0$ and it is the zero vector. Therefore, $AA^T=0\iff A=0$.

However, if $A$ is complex, the situation becomes quite different. It is no longer required that $A=0$ for $AA^T=0$. Thanks to Hanno for this case, which I left out in my original answer.

YiFan Tey
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    The OP unfortunately doesn't specify if real or complex matrices are considered. This makes a notable difference: Assume that $A^T$ denotes transposition of $\mathbf A$ without complex conjugation which is fairly standard (as notational assumption; whereas the conjugate transpose, or the "(Hermitian) adjoint" is usually denoted by $A^$). Having stated this, more can be said: In the real case $AA^T=0\iff A=0,$, cf the answer by R.C.Cowsik*. But if complex entries (continued) – Hanno Oct 27 '18 at 14:25
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    are admitted for $A$, then it gets false, an example is $AA^T=\left(\begin{smallmatrix}1&i\0&0\end{smallmatrix}\right)$$\left(\begin{smallmatrix}1&0\i&0\end{smallmatrix}\right) =\ldots$ – Hanno Oct 27 '18 at 14:25
  • Next, someone can answer for finite fields. – GEdgar Oct 27 '18 at 14:30
  • @Hanno thanks for the observation. I have edited my answer to reflect the complex case :) – YiFan Tey Oct 27 '18 at 16:19
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If the rows of $A$ are $R_1, R_2,\dots,R_n$ then the $ij^{th}$ entry of $AA^t$ is given by the dot product $R_i.R_j$.

Shweta Aggrawal
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For a real matrix $A,$ with $AA^T = 0,$ each row vector is orthogonal to all the other vectors, and in particular to itself.

So $A = 0$.

amWhy
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R.C.Cowsik
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